Question

P and Q are two points on the hyperbola $xy=c^2$ such that abscissa of P is equal to the ordinate of Q. Prove that the normals to the hyperbola at P and Q intersect on a straight line. $(ct, c/t) \rightarrow P$.

Answer 1

y=x

Answer 2

Let the equation of the hyperbola be $xy = c^2$. Let P and Q be two points on the hyperbola. Let the coordinates of P be $(x_1, y_1)$ and the coordinates of Q be $(x_2, y_2)$. Since P and Q lie on the hyperbola, we have $x_1 y_1 = c^2$ and $x_2 y_2 = c^2$. We are given that the abscissa of P is equal to the ordinate of Q, i.e., $x_1 = y_2$.

We can use the parametric form for points on the hyperbola $xy=c^2$. A point on this hyperbola can be represented as $(ct, c/t)$ for some parameter $t$. Let the coordinates of P be $(ct_1, c/t_1)$. So, $x_1 = ct_1$ and $y_1 = c/t_1$. Let the coordinates of Q be $(ct_2, c/t_2)$. So, $x_2 = ct_2$ and $y_2 = c/t_2$. The condition $x_1 = y_2$ gives $ct_1 = c/t_2$, which implies $t_1 t_2 = 1$. If we take the parameter for P as $t$, i.e., P is $(ct, c/t)$, then the parameter for Q must be $1/t$, so Q is $(c/t, c/(1/t)) = (c/t, ct)$.

Now, we find the equation of the normal to the hyperbola $xy=c^2$ at a point $(x_0, y_0)$. Differentiating $xy = c^2$ implicitly with respect to $x$, we get $y + x \frac{dy}{dx} = 0$. The slope of the tangent at $(x_0, y_0)$ is $m_t = \frac{dy}{dx}\bigg|_{(x_0, y_0)} = -\frac{y_0}{x_0}$. The slope of the normal at $(x_0, y_0)$ is $m_n = -\frac{1}{m_t} = -\frac{1}{(-y_0/x_0)} = \frac{x_0}{y_0}$. The equation of the normal at $(x_0, y_0)$ is $y - y_0 = \frac{x_0}{y_0}(x - x_0)$. Multiplying by $y_0$, we get $y y_0 - y_0^2 = x x_0 - x_0^2$, which can be rearranged as $x x_0 - y y_0 = x_0^2 - y_0^2$.

Now, we write the equation of the normal at P $(ct, c/t)$. Here $x_0 = ct$ and $y_0 = c/t$. The equation of the normal at P is: $x(ct) - y(c/t) = (ct)^2 - (c/t)^2$ $ctx - \frac{c}{t}y = c^2 t^2 - \frac{c^2}{t^2}$ Dividing by $c$ (assuming $c \neq 0$), we get: $tx - \frac{1}{t}y = ct^2 - \frac{c}{t^2}$ Multiplying by $t$, we get: $t^2 x - y = ct^3 - \frac{c}{t}$ (Equation of Normal at P)

Next, we write the equation of the normal at Q $(c/t, ct)$. Here $x_0 = c/t$ and $y_0 = ct$. The equation of the normal at Q is: $x(c/t) - y(ct) = (c/t)^2 - (ct)^2$ $\frac{c}{t}x - cty = \frac{c^2}{t^2} - c^2 t^2$ Dividing by $c$, we get: $\frac{1}{t}x - ty = \frac{c}{t^2} - ct^2$ Multiplying by $t$, we get: $x - t^2 y = \frac{c}{t} - ct^3$ (Equation of Normal at Q)

Let $(X, Y)$ be the point of intersection of the two normals. The coordinates $(X, Y)$ must satisfy both normal equations:

1. $t^2 X - Y = ct^3 - \frac{c}{t}$
2. $X - t^2 Y = \frac{c}{t} - ct^3$

We want to find the locus of $(X, Y)$ by eliminating the parameter $t$. Notice the right-hand sides of the two equations are opposite in sign: $(ct^3 - c/t) = -(c/t - ct^3)$. Let $RHS_1 = ct^3 - c/t$ and $RHS_2 = c/t - ct^3$. So $RHS_1 = -RHS_2$. The equations are:

1. $t^2 X - Y = RHS_1$
2. $X - t^2 Y = RHS_2 = -RHS_1$

Add the two equations: $(t^2 X - Y) + (X - t^2 Y) = RHS_1 + (-RHS_1)$ $t^2 X + X - Y - t^2 Y = 0$ $X(t^2 + 1) - Y(t^2 + 1) = 0$ $(X - Y)(t^2 + 1) = 0$

Since $t$ is a real parameter, $t^2 \ge 0$, which means $t^2 + 1 \ge 1$. Thus, $t^2 + 1$ is never zero. Therefore, for the product $(X - Y)(t^2 + 1)$ to be zero, we must have $X - Y = 0$.

The equation $X - Y = 0$ is the relationship between the coordinates of the intersection point $(X, Y)$. This equation represents a straight line $y=x$. Thus, the normals to the hyperbola at P and Q intersect on the straight line $y=x$.

The final answer is $\boxed{y=x}$.

Explanation of the solution:

1. Represent points P and Q on the hyperbola $xy=c^2$ parametrically as P$(ct, c/t)$ and Q$(c/t, ct)$ using the given condition.
2. Find the general equation of the normal to $xy=c^2$ at a point $(x_0, y_0)$, which is $xx_0 - yy_0 = x_0^2 - y_0^2$.
3. Write the equations of the normal at P and Q using their parametric coordinates. Normal at P: $t^2 x - y = ct^3 - c/t$. Normal at Q: $x - t^2 y = c/t - ct^3$.
4. Let $(X, Y)$ be the intersection point. The coordinates $(X, Y)$ satisfy both normal equations.
5. Add the two equations for $(X, Y)$: $(t^2 X - Y) + (X - t^2 Y) = (ct^3 - c/t) + (c/t - ct^3)$.
6. Simplify the sum: $X(t^2+1) - Y(t^2+1) = 0$, which leads to $(X-Y)(t^2+1) = 0$.
7. Since $t^2+1 \neq 0$ for real $t$, we must have $X-Y=0$.
8. The locus of the intersection point $(X, Y)$ is the straight line $x-y=0$ or $y=x$.

Answer: The normals to the hyperbola at P and Q intersect on the straight line $y=x.